Radioactive decay is a first order process. Radioactive carbon in wood sample decays with a half-life of 5770 yr. What is the rate constant (in yr-1) for the decay? What fraction would remain after 11540 yr?

Question

Radioactive decay is a first order process. Radioactive carbon in wood sample decays with a half-life of 5770 yr. What is the rate constant (in yr-1) for the decay? What fraction would remain after 11540 yr? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

k=( ln 2/t1/2)=(0.693/5770)yr-1=1.2×10-4yr-1 Also kt= ln(1/f)=( ln 2/5770)×11540= ln 4 ⇒ f=(1/4)=0.25

Q. Radioactive decay is a first order process. Radioactive carbon in wood sample decays with a half-life of $5770 yr$ . What is the rate constant (in $y r^{- 1})$ for the decay? What fraction would remain after $11540 yr$ ?

$k = \frac{l n 2}{t _{1/2}} = \frac{0.693}{5770} y r^{- 1} = 1.2 \times 1 0^{- 4} y r^{- 1}$
Also $k t = ln \frac{1}{f} = \frac{l n 2}{5770} \times 11540 = ln 4$
$\Rightarrow f = \frac{1}{4} = 0.25$

Q. Radioactive decay is a first order process. Radioactive carbon in wood sample decays with a half-life of 5770yr. What is the rate constant (in yr−1) for the decay? What fraction would remain after 11540yr?

k=t1/2​ln2​=57700.693​yr−1=1.2×10−4yr−1 Also kt=lnf1​=5770ln2​×11540=ln4 ⇒f=41​=0.25

Q. Radioactive decay is a first order process. Radioactive carbon in wood sample decays with a half-life of $5770 yr$ . What is the rate constant (in $y r^{- 1})$ for the decay? What fraction would remain after $11540 yr$ ?

$k = \frac{l n 2}{t _{1/2}} = \frac{0.693}{5770} y r^{- 1} = 1.2 \times 1 0^{- 4} y r^{- 1}$
Also $k t = ln \frac{1}{f} = \frac{l n 2}{5770} \times 11540 = ln 4$
$\Rightarrow f = \frac{1}{4} = 0.25$